# Complex numbers problem

1. Mar 7, 2006

### meee

need a lil help here

thnx

Last edited: Mar 7, 2006
2. Mar 7, 2006

### benorin

Set g(z)=z, that is, write 2 + zi = z and then solve for z.

3. Mar 7, 2006

### meee

ok.. thanks, any tips in solving for z?

i got.. z = 2 + zi
z/z = 2/z + i
1 = 2/z + i
2/z = 1 - i
z= 2/(1-i)

dont think im going the right way

Last edited: Mar 7, 2006
4. Mar 7, 2006

### benorin

$$2 + iz = z$$
$$2 = z -iz = z(1-i)$$

and so, like you have

$$z= \frac{2}{1-i}$$

now multiply by the conjugate of the denominator on top and bottom

5. Mar 7, 2006

### meee

wel..... we can divide 2 by (1 - i) to get z...

OHHHhhhhhhhhhhhhhhhhhhhh ahhhhhhhhhh.!!!1 forgot that!1 multiple by conjugate... how silly of me

thankyou so much!!!! !!!!!!1

yaaay.....

bottom: (1-i) * (1+i) = 2
top: 2*(1+i) = 2+2i

= (2+2i)/2
= 1+i

Yayaya Thnnx So mUcH!

Last edited: Mar 7, 2006
6. Mar 7, 2006

### benorin

$$z= \frac{2}{1-i} =\left( \frac{2}{1-i}\right) \left( \frac{1+i}{1+i} \right) = \frac{2(1+i)}{(1-i)(1+i)} = \frac{2(1+i)}{1^2-i^2} = \frac{2(1+i)}{1-(-1)} = 1+i$$

7. Mar 7, 2006

### meee

nice diagram!

tankyou somuch